When you think about the limitation, lim_(n->oo) a_n, don't forget that
oo is NOT a infinitely large color(blue)"number", BUT a color(red)"situation" n is increasing infinitely.
[color(red)"example1"] Where does n^2/n go when n goes to infinity?
Here we must evaluate lim_(n->oo)n^2/n
It is true lim_(n->oo)n=oo and lim_(n->oo)n^2=oo, but if you think
lim_(n->oo)n^2/n=oo/oo=1 you are wrong.
The situation shows n^2 is growing much faster than n and thus the limitation cannot be 1.
To solve this, you first need to color(red)"reduce the fraction" and then let n->oo.
lim_(n->oo)n^2/n=color(red)(lim_(n->oo)n/1=lim_(n->oo)n=oo)
is the correct way.
[color(blue)"example2"] What is lim_(n->oo)(n-sqrt(n^2+2n))?
You might think:
lim_(n->oo)n=oo and lim_(n->oo)sqrt(n^2+2n)=oo and oo-oo=0・・・Just wait! This is wrong again.
Here we should color(blue)"ratonalize" n-sqrt(n^2+2n).
lim_(n->oo)(n-sqrt(n^2+2n))=lim_(n->oo)((n-sqrt(n^2+2n))(n+sqrt(n^2+2n)))/(n+sqrt(n^2+2n))
=color(red)(lim_(n->oo)(-2n)/(n+sqrt(n^2+2n))) ・・(A)
(A) is oo/oo again, so divide both numerator and denominator by n.
(A)=color(blue)(-lim_(n->oo)2/(1+sqrt(1+2/n)))
=-2/(1+1)=-1 and this is the goal.
In general, oo/oo, 0/0, ooxx0, oo-oo, 1^oo or oo^0 typed limitations are called color(green)"indeterminate form" and they cannot be calculated like numbers. They should be treated carefully.
